Help.............Triangle ABC is a right triangle with angle ACB as the right angle. Angle ABC =60 degrees and AB=10. Let P be randomly chosen inside triangle ABC and extend BP to meet AC at D.
What is the probability that BD>5*sqrt(2) Help.............Triangle ABC is a right triangle with angle ACB as the right angle. Angle ABC =60 degrees and AB=10. Let P be randomly chosen inside triangle ABC and extend BP to meet AC at D.
What is the probability that BD>5*sqrt(2) @Mathematics
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how did u take as 15 and 45?
i never said that CBD is a right isosceles triangle
since we want hat BD> 5sqrt(2)........(I)
we know that....10/sine(BDA) = BD/ sine(DAB)
since.... sine(DAB) = 1/2
therefore..... BD = 5/ sine(BDA)
put in equation no....(I)
we get..... 5/sine(BDA)> 5*sqrt(2)
yaha tak smjh me aa gaya kya sb kuch????
so we get............ sin (BAD)<1/sqrt(2)
or 180degre< BAD<135degree
got it or not????
From here i understood how to solve
Thankx very very much fr ure help.......
now do u understand the figure that i have made....its the boundary line of the probability...that means in that figure if we move d to the right then we will get the desired probability.......
And onee day hopefully like u i will be an IITIAN
Yeah i understood
THANX A LOT.......
listen tio the important fact........ur answer would be now.......
area of triangle(CBD)/ area oof the triangle (DBA)
this is the main thing....
bye.....i'm leaving now...mujhe khana khane jana h aur bhi kuch kam h....i'll be back in two hrs now...do send me a request on facebook.......